Question

Cho a,b,c >0 và \(\frac{a+b}{3}\)=\(\frac{b+c}{4}\)=\(\frac{c+a}{5}\)

Tính giá trị biểu thức M=10a+b-7c+2017

Answer 1

Ta có: 

\(\hept{\begin{cases}\frac{a+b}{3}=\frac{b+c}{4}\Rightarrow4a+4b=3b+3c\Rightarrow4a+b-3c=0\left(1\right)\\\frac{b+c}{4}=\frac{c+a}{5}\Rightarrow5b+5c=4c+4a\Rightarrow4a-5b-c=0\Rightarrow4a=5b+c\left(2\right)\\\frac{c+a}{5}=\frac{a+b}{3}\Rightarrow3c+3a=5a+5b\Rightarrow2a+5b-3c=0\Rightarrow3c=2a+5b\left(3\right)\end{cases}}\)

Thay (2) vào (1) ta có: 3b=c

Thay (3) và (1) ta có: 2b=a

Vậy M=10a+b-7c+2017=10.2b+b-7.3b+2017=21b-21b+2017=0+2017=2017